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\title{\heiti\zihao{2} 习题8.3}
\author{中书君}
\date{\songti 2021年1月13日}

\begin{document}
\maketitle
\section{求下列曲线的弧长:}
\subsection{旋轮线 $\left\{\begin{array}{l}x=a(t-\sin t) \\ y=a(1-\cos t)\end{array}, 0 \leqslant t \leqslant 2 \pi, a>0\right.$}
\textbf{解}\quad
$$
\begin{aligned}
L&=\int_{0}^{2 \pi} a \sqrt{(1-\cos t)^{2}+\sin ^{2} t} d t
\\&=a \int_{0}^{2 \pi} \sqrt{1+\sin ^{2} t+\cos ^{2}t-2 \cos t} d t=\sqrt{2} a \int_{0}^{2 \pi} \sqrt{1-\cos t} d t
\\&=\sqrt{2} a \int_{0}^{2 \pi} \sqrt{2} \sin \frac{t}{2} d t\\&
\\&=2 a \int_{0}^{2 \pi} \sin \frac{t}{2} d t=8 a
\end{aligned}
$$

\subsection{星形线 $\left\{\begin{array}{l}x=a \cos ^{3} t \\ y=a \sin ^{3} t\end{array}, 0 \leqslant t \leqslant 2 \pi, a>0\right.$}
\textbf{解}\quad
星形线 $\left\{\begin{array}{l}x=a \cos ^{3} t \\ y=a \sin { }^{3} t\end{array}, 0 \leqslant t \leqslant 2 \pi, a>0\right.$
$$
\begin{aligned}
L &=4 a \int_{0}^{\frac{\pi}{2}} \sqrt{9 \sin ^{4} x \cos ^{2} x+9 \cos ^{4} x \sin ^{2} x} \\
&=12 a \int_{0}^{\frac{\pi}{2}} \sin x \cos x d x \\
&=6 a \int_{0}^{\frac{\pi}{2}} \sin 2 x d x \\
&=6 a
\end{aligned}
$$

\subsection{曲线 $y=\ln \cos x, x \in\left[0, \frac{\pi}{3}\right]$}
\textbf{解}\quad
曲线 $y=\ln \cos x, x \in\left[0, \frac{\pi}{3}\right]$
$$
\begin{aligned}
L &=\int_{0}^{\frac{\pi}{3}} \sqrt{1+\tan ^{2} x} d x \\
&=\int_{0}^{\frac{\pi}{3}} \sec x d x
=\ln \left(\tan \frac{\pi}{3}+\sec \frac{\pi}{3}\right)\\&=\ln (2+\sqrt{3})
\end{aligned}
$$

\subsection{$y=x^{\frac{3}{2}}, 0 \leqslant x \leqslant 4$}
\textbf{解}\quad
$y=x^{\frac{3}{2}}, 0 \leqslant x \leqslant 4$
$$
L=\int_{0}^{4} \sqrt{1+\frac{9}{4} x} \mathrm{~d} x=\frac{80 \sqrt{10}-8}{27}
$$

\subsection{心形线 $r=a(1+\cos \theta), a>0$}
\textbf{解}\quad
心形线 $r=a(1+\cos \theta), a>0 .$
$$
\begin{aligned}
L &=2a \int_{0}^{ \pi} \sqrt{1+\cos ^{2} \theta+2 \cos \theta+\sin ^{2} \theta} d \theta=2a \int_{0}^{ \pi} \sqrt{2+2 \cos \theta} d \theta \\
&=2\sqrt{2} a \int_{0}^{ \pi} \sqrt{1+2 \cos ^{2} \frac{\theta}{2}-1} d \theta=4 a \int_{0}^{2 \pi} \cos \frac{\theta}{2} d \theta\\&
=8a
\end{aligned}
$$

\section{求下列曲线在指定点的曲率：}
\subsection{$x y=1,$ 在点 $(1,1)$}
\textbf{解}\quad
$$
y=\frac{1}{x}, y^{\prime}=-x^{-2}, y^{\prime \prime}=2 x^{-3}
$$

将$x=1$带入上述式子，并将$y$带入公式$(1)$可得$k=\frac{\sqrt{2}}{2}$

\subsection{摆线 $\left\{\begin{array}{l}x=a(t-\sin t), \\ y=a(1-\cos t)\end{array} a>0,\right.$ 在参数 $t=\frac{\pi}{2}$ 所对应的点}
\textbf{解}\quad
$$
x^{\prime}(t)=a(1-\cos t), y^{\prime}(t)=a \sin t, x^{\prime \prime}(t)=a \sin t, y^{\prime \prime}(t)=a \cos t
$$

将$t=\frac{\pi}{2}$带入并带入公式$(2)$可得$k=\frac{1}{2 \sqrt{2} a}$

\section{求下列曲线的曲率与曲率半径:}
\subsection{抛物线 $x^{2}=2 p y,$ 其中 $p>0$}
\textbf{解}\quad
$$
\frac{\mathrm{d} x}{\mathrm{~d} y}=\frac{y}{p}, \frac{\mathrm{d}^{2} x}{\mathrm{~d} y^{2}}=\frac{1}{p}, K=\frac{\sqrt{p}}{(p+2 x)^{\frac{3}{2}}}, R=\frac{1}{K}=\frac{(p+2 x)^{\frac{3}{2}}}{\sqrt{p}}
$$

\subsection{星形线 $\left\{\begin{array}{l}x=a \cos ^{3} t, \\ y=a \sin { }^{3} t\end{array} 0 \leqslant t \leqslant 2 \pi, a>0\right.$}
\textbf{解}\quad
$$
\begin{array}{l}
\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3 a \sin ^{2} t \cos t}{-3 a \cos ^{2} t \sin t}=-\tan t, \quad \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=\frac{1}{3 a} \cdot \frac{1}{\cos ^{4} t \sin t}, \\
K=\frac{1}{3 a} \cdot \frac{1}{|\sin t \cos t|}=\frac{1}{3 \sqrt[3]{|a x y|}}, \quad R=3 \sqrt[3]{|a x y|}
\end{array}
$$

\subsection{双纽线 $\left(x^{2}+y^{2}\right)^{2}=a^{2}\left(x^{2}-y^{2}\right),$ 其中 $a>0$}
\textbf{解}\quad
用极坐标形式表示双纽线：$x=r \cos \theta,y = r \sin \theta$

即可得$r^{2}=a^{2}\cos 2\theta$
$$
\begin{aligned}
&r^{2}+2 r^{\prime 2}-r r^{\prime \prime}=3 r^{2}+3 r^{\prime 2}\\&
r^{2}+r^{\prime 2}=\frac{4 a^{4}}{r^{2}}\\&
K=\frac{3 r}{2 a^{2}},R=\frac{1}{K}
\end{aligned}
$$

\subsection{悬链线 $y=\frac{a}{2}\left(\mathrm{e}^{\frac{x}{a}}+\mathrm{e}^{-\frac{x}{a}}\right),(a>0)$}
\textbf{解}\quad
$$
\begin{array}{c}
y^{\prime}=\frac{1}{2}\left(\mathrm{e}^{\frac{x}{a}}-\mathrm{e}^{-\frac{x}{a}}\right), y^{\prime \prime}=\frac{1}{2 a}\left(\mathrm{e}^{\frac{x}{a}}+\mathrm{e}^{-\frac{x}{a}}\right)=\frac{y}{a^{2}}, \text { 由 } \mathrm{于} \sqrt{1+y^{\prime}}=\frac{y}{a} \\
K=\frac{a}{y^{2}}, \quad R=\frac{1}{K}=\frac{y^{2}}{a}
\end{array}
$$








\end{document}
\subsection{}
\textbf{解}\quad

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\textbf{证}\quad

\textbf{\textcolor{red}{注}}\quad